Suppose a rock is thrown upwards from the top of a building 200 feet high with an initial velocity of 96 feet per second. The height (in feet) of the rock above the ground t seconds after being released is given by the formula

h(t) = -16t^2 + 96t + 200

1.When will it hit the ground?
2.How high will it be after 2.1 seconds?
3.What is the maximum height?

Question

Suppose a rock is thrown upwards from the top of a building 200 feet high with an initial velocity of 96 feet per second. The height (in feet) of the rock above the ground t seconds after being released is given by the formula

h(t) = -16t^2 + 96t + 200

1.When will it hit the ground? 
 2.How high will it be after 2.1 seconds? 
 3.What is the maximum height?

anonymous · Answer

1. It will hit the ground when h(t) = 0 so you need to solve
$$-16t^2+96t+200=0$$

anonymous · Answer

try the quadratic formula

kj4uts · Answer

Would it be t=3-√ 43/2=-1.63 and 3+√ 43/2=7.63

anonymous · Answer

yes, and since time can't be negative, it hits the ground at 7.63 seconds.

anonymous · Answer

For 2, plug in 2.1 for t to find h(2.1).

kj4uts · Answer

331.04

anonymous · Answer

yes. and then for 3 there are two steps.
Use $t=-\frac{ b }{ 2a }$ to find the time of max height.
Then plug that time into the h(t) equation

kj4uts · Answer

Im a little confused as to what I plug into b and a?

anonymous · Answer

the same values you used in the quadratic equation
a = -16, b = 96, c = 200

anonymous · Answer

$$t=-\frac{ 96 }{ 2(-16) }$$

kj4uts · Answer

3

anonymous · Answer

now plug 3 in for t in the original equation

kj4uts · Answer

Ok I got 344 thank you for explaining this step by step :)

anonymous · Answer

yeah that's it. you're welcome. :)